Get thmtools style with mdframed (issues with the title and MWE provided)

Question

I am trying to achieve the following output with mdframed, but I can't. I did the proof in the attached picture below with thmtools, but I am abandoning it because some of my proofs are very long and I want them to be split between pages to avoid getting to large white spaces at the end of some pages (something that apparently thmtools cannot handle easily). What I can get is this:

But the output I am trying to replicate should look as follows:

As you can see, what I need is:

(a) The title in the same line as the text.

(b) The word "Proof" before the name of the theorem being proven.

(c) A long horizontal dash between Proof and the title of the theorem being proven.

(d) A colon after the title of the theorem being proven.

Does it make sense? Here you can find a MWE that replicates what I get (except the font and the color of the math; because those are irrelevant to my question and would only make a code more complex without need).

\documentclass[a4paper]{report}
\pagestyle{plain}

\usepackage[dvipsnames]{xcolor}
\usepackage{amsmath, mathtools, amsthm, mathrsfs, amssymb}
\usepackage{mdframed}
\let\proof\relax
\let\endproof\relax
\newmdenv[linecolor=Gray,frametitle=Proof]{proof}

\begin{document}
\begin{proof}[frametitle={\textbf{von-Neumann--Morgenstern Expected Utility Theorem I}}]
Let's prove that if there exists $U: \mathcal{L}(X) \longrightarrow \mathbb{R}$, a preference $\succsim$ on $\mathcal{L}(X)$ satisfies X and Y. Assume there exists a utility function $U: \mathcal{L}(X) \longrightarrow \mathbb{R}$ representing $\succsim$ such that $U$ satisfies the expected utility property.
\begin{enumerate}
    \item To show that $\succsim$ \textbf{is continuous}, let $x,y,z \in \mathcal{L}(X)$ such that $x \succ y \succ z$. Since $U$ represents $\succsim$, $U(x) > U(y) > U(z)$. The set of real numbers is convex; and hence there exists $p \in (0,1)$ such that $p \cdot U(x) + (1-p) \cdot U(z) = U(y)$. By the expected utility property, $U(p \odot x \oplus (1-p) \odot z) = p \cdot U(x) + (1-p) \cdot U(z) = U(y)$. Since $U$ represents $\succsim$, $p \odot x \oplus (1-p) \odot z \sim y$. Now, let $q,r \in [0,1]$ be such that $q >p>r$. Then:
    \begin{flalign*}
        && q \cdot U(x) + (1-q) \cdot U(z) & > p \cdot U(x) + (1-p) \cdot U(z)\\
        && &> r \cdot U(x) + (1-r)\cdot U(z)
    \end{flalign*}

    And by the expected utility property, and the hypothesis that $U$ represents $\succsim$, it follows that $q \odot x \oplus (1-q) \odot z \succ y \succ r \odot x \oplus (1-r) \odot z$.
    \item To show that $\succsim$ \textbf{satisfies independence}, let $x,y,z \in \mathcal{L}(X)$ and let $p \in [0,1]$. Since $U$ represents $\succsim$, $x \succ y \Longleftrightarrow U(x) > U(y)$. Hence, $x \succ y \Longleftrightarrow p \cdot U(x) + (1-p) \cdot U(z) > p \cdot U(y) + (1-p) \cdot U(z)$. Since $U$ satisfies the expected utility property, $x \succ y \Longleftrightarrow [p \odot x \oplus (1-p) \odot z] \succ [p \odot y \oplus (1-p) \odot z]$. Since $U$ represents $\succsim$, $x \sim y \Longleftrightarrow U(X) = U(y)$. Hence, $x \sim y \Longleftrightarrow p \cdot U(x) + (1-p) \cdot U(z) = p \cdot U(y) + (1-p) \cdot U(z)$. Since $U$ satisfies the expected utility property, $x \sim y \Longleftrightarrow [p \odot x \oplus (1-p) \odot z] \sim [p \odot y \oplus (1-p) \odot z]$.
\end{enumerate}
\end{proof}

\end{document}

Could anybody help me, please? Thank you all in advance for your time.

Answer 1

With a bit of help from https://tex.stackexchange.com/a/185168/36296:

\documentclass[a4paper]{report}

\usepackage{amsmath, amssymb, amsthm}
\usepackage{mdframed}

\newtheoremstyle{mystyle}%                % Name
  {0pt}%                                     % Space above
  {}%                                     % Space below
  {\itshape}%                             % Body font
  {}%                                     % Indent amount
  {\bfseries}%                            % Theorem head font
  {:}%                                    % Punctuation after theorem head
  { }%                                    % Space after theorem head, ' ', or \newline
  {\thmname{#1} -- \thmnote{#3}}%       % Theorem head spec (can be left empty, meaning `normal')
\theoremstyle{mystyle}

\let\proof\relax
\let\endproof\relax

\newmdtheoremenv[innerleftmargin=0.1cm,innerrightmargin=0.1cm,innertopmargin=0.1cm,innerbottommargin=0.1cm]{proof}{Proof}

\AtEndEnvironment{proof}{\hfill$\square$}%

\begin{document}
\begin{proof}[von-Neumann--Morgenstern Expected Utility Theorem I]
Let's prove that if there exists $U: \mathcal{L}(X) \longrightarrow \mathbb{R}$, a preference $\succsim$ on $\mathcal{L}(X)$ satisfies X and Y. Assume there exists a utility function $U: \mathcal{L}(X) \longrightarrow \mathbb{R}$ representing $\succsim$ such that $U$ satisfies the expected utility property.
\begin{enumerate}
    \item To show that $\succsim$ \textbf{is continuous}, let $x,y,z \in \mathcal{L}(X)$ such that $x \succ y \succ z$. Since $U$ represents $\succsim$, $U(x) > U(y) > U(z)$. The set of real numbers is convex; and hence there exists $p \in (0,1)$ such that $p \cdot U(x) + (1-p) \cdot U(z) = U(y)$. By the expected utility property, $U(p \odot x \oplus (1-p) \odot z) = p \cdot U(x) + (1-p) \cdot U(z) = U(y)$. Since $U$ represents $\succsim$, $p \odot x \oplus (1-p) \odot z \sim y$. Now, let $q,r \in [0,1]$ be such that $q >p>r$. Then:
    \begin{flalign*}
        && q \cdot U(x) + (1-q) \cdot U(z) & > p \cdot U(x) + (1-p) \cdot U(z)\\
        && &> r \cdot U(x) + (1-r)\cdot U(z)
    \end{flalign*}

    And by the expected utility property, and the hypothesis that $U$ represents $\succsim$, it follows that $q \odot x \oplus (1-q) \odot z \succ y \succ r \odot x \oplus (1-r) \odot z$.
    \item To show that $\succsim$ \textbf{satisfies independence}, let $x,y,z \in \mathcal{L}(X)$ and let $p \in [0,1]$. Since $U$ represents $\succsim$, $x \succ y \Longleftrightarrow U(x) > U(y)$. Hence, $x \succ y \Longleftrightarrow p \cdot U(x) + (1-p) \cdot U(z) > p \cdot U(y) + (1-p) \cdot U(z)$. Since $U$ satisfies the expected utility property, $x \succ y \Longleftrightarrow [p \odot x \oplus (1-p) \odot z] \succ [p \odot y \oplus (1-p) \odot z]$. Since $U$ represents $\succsim$, $x \sim y \Longleftrightarrow U(X) = U(y)$. Hence, $x \sim y \Longleftrightarrow p \cdot U(x) + (1-p) \cdot U(z) = p \cdot U(y) + (1-p) \cdot U(z)$. Since $U$ satisfies the expected utility property, $x \sim y \Longleftrightarrow [p \odot x \oplus (1-p) \odot z] \sim [p \odot y \oplus (1-p) \odot z]$.
\end{enumerate}
\end{proof}

\end{document}

Answer 2

I propose a solution with the simpler framed environment option of ntheorem. One advantage is the automatic (and correct) placement of the qed symbol, even when the proof ends in a displayed equation. It is activated by the thmmarks option.

I redefined the nonumberplain predefined style under the name myproof so as to incorporate the emdash between the name of the environment and the optional argument, in the place of the pair of parentheses.

Also I obtain a better layout of the enumerate environment inside the frame with enumitem.

\documentclass[a4paper]{report}
\pagestyle{plain}
\usepackage{geometry}%
 \usepackage[dvipsnames]{xcolor}
\usepackage{amsmath, mathtools, mathrsfs, amssymb}
\usepackage{framed, enumitem} %
\usepackage[framed, thref, amsmath, thmmarks]{ntheorem} %
\makeatletter
\newtheoremstyle{myproof}%
{\item[\theorem@headerfont\hskip\labelsep ##1\theorem@separator]}%
{\item[\theorem@headerfont\hskip \labelsep ##1~---~##3\theorem@separator]}
\makeatother
\theoremheaderfont{\bfseries}
\theorembodyfont{\mdseries}
\theoremseparator{:}
\theoremsymbol{\ensuremath{\square}}
\theoremstyle{myproof}
\newframedtheorem{proof}{Proof}

\begin{document}

\begin{proof}[von\,Neumann-Morgenstern Expected Utility Theorem I]
Let's prove that if there exists $U: \mathcal{L}(X) \longrightarrow \mathbb{R}$, a preference $\succsim$ on $\mathcal{L}(X)$ satisfies X and Y. Assume there exists a utility function $U: \mathcal{L}(X) \longrightarrow \mathbb{R}$ representing $\succsim$ such that $U$ satisfies the expected utility property.
\begin{enumerate}[wide=0pt, leftmargin=*]
    \item To show that $\succsim$ \textbf{is continuous}, let $x,y,z \in \mathcal{L}(X)$ such that $x \succ y \succ z$. Since $U$ represents $\succsim$, $U(x) > U(y) > U(z)$. The set of real numbers is convex; and hence there exists $p \in (0,1)$ such that $p \cdot U(x) + (1-p) \cdot U(z) = U(y)$. By the expected utility property, $U(p \odot x \oplus (1-p) \odot z) = p \cdot U(x) + (1-p) \cdot U(z) = U(y)$. Since $U$ represents $\succsim$, $p \odot x \oplus (1-p) \odot z \sim y$. Now, let $q,r \in [0,1]$ be such that $q >p>r$. Then:
    \begin{flalign*}
        && q \cdot U(x) + (1-q) \cdot U(z) & > p \cdot U(x) + (1-p) \cdot U(z)\\
        && &> r \cdot U(x) + (1-r)\cdot U(z)
    \end{flalign*}

    And by the expected utility property, and the hypothesis that $U$ represents $\succsim$, it follows that $q \odot x \oplus (1-q) \odot z \succ y \succ r \odot x \oplus (1-r) \odot z$.
    \item To show that $\succsim$ \textbf{satisfies independence}, let $x,y,z \in \mathcal{L}(X)$ and let $p \in [0,1]$. Since $U$ represents $\succsim$, $x \succ y \Longleftrightarrow U(x) > U(y)$. Hence, $x \succ y \Longleftrightarrow p \cdot U(x) + (1-p) \cdot U(z) > p \cdot U(y) + (1-p) \cdot U(z)$. Since $U$ satisfies the expected utility property, $x \succ y \Longleftrightarrow [p \odot x \oplus (1-p) \odot z] \succ [p \odot y \oplus (1-p) \odot z]$. Since $U$ represents $\succsim$, $x \sim y \Longleftrightarrow U(X) = U(y)$. Hence, $x \sim y \Longleftrightarrow p \cdot U(x) + (1-p) \cdot U(z) = p \cdot U(y) + (1-p) \cdot U(z)$. Since $U$ satisfies the expected utility property, $x \sim y \Longleftrightarrow [p \odot x \oplus (1-p) \odot z] \sim [p \odot y \oplus (1-p) \odot z]$.
\end{enumerate} %
\end{proof}

\end{document}